The equation of a circle $C$ is $x^2+y^2+2x+16y+49 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2+2x) + (y^2+16y) = -49$ $(x^2+2x+1) + (y^2+16y+64) = -49 + 1 + 64$ $(x+1)^{2} + (y+8)^{2} = 16 = 4^2$ Thus, $(h, k) = (-1, -8)$ and $r = 4$.